Sequences and Their Limits

Consider the sequence $\left\{{\frac {1}{n}}\right\}_{n=1}^{\infty }=\left\{1,{\frac {1}{2}},{\frac {1}{3}},{\frac {1}{4}},...\right\}$. As $n\to \infty$, it appears as though ${\frac {1}{n}}\to 0$. In fact, we know that this is true since $\lim _{n\to \infty }{\frac {1}{n}}=0$. We will now formalize the definition of a limit with regards to sequences.

Definition: If $\{a_{n}\}$ is a sequence, then $\lim _{n\to \infty }a_{n}=L$ means that for every $\epsilon >0$ there exists a corresponding $N\in \mathbb {N}$ such that if $n\geq N$, then $\mid a_{n}-L\mid <\epsilon$. If this limit exists, then we say that the sequence $\{a_{n}\}$ Converges, and if this limit doesn't exist then we say say the sequence $\{a_{n}\}$ Diverges.

We note that our definition of the limit of a sequence is very similar to the limit of a function, in fact, we can think of a sequence as a function whose domain is the set of natural numbers $\mathbb {N}$. From this notion, we obtain the very important theorem:

Theorem 1: If $\{a_{n}\}$ is a sequence and a function $f(x)$, then if $f(n)=a_{n}$ where $n\in \mathbb {N}$ and $\lim _{x\to \infty }f(x)=L$, then $\lim _{n\to \infty }f(n)=L$.

• Proof of Theorem 1: Let $\epsilon >0$ be given. We know that $\lim _{x\to \infty }f(x)=L$ which implies that $\forall \epsilon >0\exists N\in \mathbb {R}$ such that if $x\geq k$ then $\mid f(x)-L\mid <\epsilon$.

• Now we want to show that $\forall \epsilon >0\exists N\in \mathbb {N}$ such that if $n\geq N$ then $\mid a_{n}-L\mid <\epsilon$. We will choose $N=\mathrm {max} \{1,\lceil k\rceil \}$. This ensures that $N$ is an integer.

• Now since $N\geq k$ then it follows that if $\forall n\geq N$ then $\mid f(n)-L\mid <\epsilon$. But $f(n)=a_{n}$ and so $\mid a_{n}-L\mid <\epsilon$ so $\lim _{n\to \infty }a_{n}=L$. $\blacksquare$

Important Note: The converse of this theorem is not implied to be true! That is if $f(n)=a_{n}$ and $\lim _{n\to \infty }f(n)=L$, then this does NOT imply that $\lim _{x\to \infty }f(x)=L$. For example, consider the sequence $\{\sin(2\pi n)\}=\{0,0,0,...\}$. Clearly this sequence converges at 0. However, the function $f(x)=\sin(2\pi x)$ does not converge, instead, it diverges as it oscillates between $-1$ and $1$.

For example, consider the sequence $\left\{{\frac {n+2}{n}}\right\}_{n=1}^{\infty }$. If we let $f(n)={\frac {n+2}{n}}$ be a function whose domain is the natural numbers, then we calculate the limit of this function like we have in the past, namely:

${\begin{aligned}\lim _{n\to \infty }f(n)=\lim _{n\to \infty }{\frac {n+2}{n}}=\lim _{n\to \infty }1+{\frac {2}{n}}=1\end{aligned}}$

Therefore the limit of our sequence $f(n)=a_{n}$ is 1, that is, $\left\{{\frac {n+2}{n}}\right\}_{n=1}^{\infty }$ converges to 1 as $n\to \infty$.

Now let's look at another major theorem.

Theorem 2: If $\lim _{n\to \infty }a_{n}=L$ and a function $f$ is continuous at $L$, then $\lim _{n\to \infty }f(a_{n})=f\left(\lim _{n\to \infty }a_{n}\right)=f(L)$.

• Proof of Theorem 2: If $f$ is a continuous function at $x=L$, then we know that $f(L)$ is defined and that $\lim _{x\to L}f(x)=f(L)$. By the definition of a limit, $\forall \epsilon >0\exists \delta >0$ such that if $0<\mid x-L\mid <\delta$ then $\mid f(x)-f(L)\mid <\epsilon$. Let $x=a_{n}$ so then if $0<\mid a_{n}-L\mid <\delta$ then $\mid f(a_{n})-f(L)\mid <\epsilon$.

• We want to show that $\lim _{n\to \infty }f(a_{n})=f(L)$, that is $\forall \epsilon >0\exists N\in \mathbb {N}$ such that if $n\geq N$ then $\mid f(a_{n})-f(L)\mid <\epsilon$, which is what we showed above.

We will now look at some important limit laws regarding sequences

Limit Laws of Convergent Sequences

We will now look at some very important limit laws regarding convergent sequences, all of which are analogous to the limit laws for functions that we already know of.

• Law 1 (Addition Law of Convergent Sequences): If the limits of the sequences $\{a_{n}\}$ and $\{b_{n}\}$ are convergent, that is $\lim _{n\to \infty }a_{n}=A$ and $\lim _{n\to \infty }b_{n}=B$ then $\lim _{n\to \infty }(a_{n}+b_{n})=\lim _{n\to \infty }a_{n}+\lim _{n\to \infty }b_{n}=A+B$.

• Proof of Law 1: Let $\{a_{n}\}$ and $\{b_{n}\}$ be convergent sequences. By the definition of a sequence being convergent, we know that $\lim _{n\to \infty }a_{n}=A$ and $\lim _{n\to \infty }b_{n}=B$ for some $A,B\in \mathbb {R}$.

• Now let $\epsilon >0$ be given, and recall that $\lim _{n\to \infty }a_{n}=A$ implies that $\forall \epsilon \exists N_{1}\in \mathbb {N}$ such that if $n\geq N_{1}$ then $\mid a_{n}-A\mid <{\frac {\epsilon }{2}}$. Similarly, $\lim _{n\to \infty }b_{n}=B$ implies that $\forall \epsilon \exists N_{2}\in \mathbb {N}$ such that if $n\geq N_{2}$ then $\mid b_{n}-B\mid <{\frac {\epsilon }{2}}$.

• We will choose $N=\mathrm {max} \{N_{1},N_{2}\}$. By choosing the larger of $N_{1}$ and $N_{2}$, we ensure that the if $n\geq \mathrm {max} \{N_{1},N_{2}\}$, then $\mid a_{n}-A\mid <{\frac {\epsilon }{2}}$ and $\mid b_{n}-B\mid <{\frac {\epsilon }{2}}$. We now want to show that if $n\geq N$, then $\mid (a_{n}+b_{n})-(A+B)\mid <\epsilon$. By the triangle inequality we obtain that:

${\begin{aligned}\quad \quad \mid (a_{n}+b_{n})-(A+B)\mid =\mid (a_{n}-A)+(b_{n}-B)\mid \leq \mid a_{n}+A\mid +\mid b_{n}+B\mid <{\frac {\epsilon }{2}}+{\frac {\epsilon }{2}}=\epsilon \end{aligned}}$

• Therefore $\lim _{n\to \infty }(a_{n}+b_{n})=A+B$. $\blacksquare$

• Law 2 (Difference Law of Convergent Sequences): If the limits of the sequences $\{a_{n}\}$ and $\{b_{n}\}$ are convergent, that is $\lim _{n\to \infty }a_{n}=A$ and $\lim _{n\to \infty }b_{n}=B$ then $\lim _{n\to \infty }(a_{n}-b_{n})=\lim _{n\to \infty }a_{n}-\lim _{n\to \infty }b_{n}=A-B$.

• Proof of Law 2: Let $\{a_{n}\}$ and $\{b_{n}\}$ be convergent sequences. By the definition of a sequence being convergent, we know that $\lim _{n\to \infty }a_{n}=A$ and $\lim _{n\to \infty }b_{n}=B$ for some $A,B\in \mathbb {R}$.

• Now let $\epsilon >0$ be given, and recall that $\lim _{n\to \infty }a_{n}=A$ implies that $\forall \epsilon \exists N_{1}\in \mathbb {N}$ such that if $n\geq N_{1}$ then $\mid a_{n}-A\mid <{\frac {\epsilon }{2}}$. Similarly, $\lim _{n\to \infty }b_{n}=B$ implies that $\forall \epsilon \exists N_{2}\in \mathbb {N}$ such that if $n\geq N_{2}$ then $\mid b_{n}-B\mid <{\frac {\epsilon }{2}}$.

• We will choose $N=\mathrm {max} \{N_{1},N_{2}\}$. By choosing the larger of $N_{1}$ and $N_{2}$, we ensure that the if $n\geq \mathrm {max} \{N_{1},N_{2}\}$, then $\mid a_{n}-A\mid <{\frac {\epsilon }{2}}$ and $\mid b_{n}-B\mid <{\frac {\epsilon }{2}}$. We now want to show that if $n\geq N$, then $\mid (a_{n}-b_{n})-(A-B)\mid <\epsilon$. By the triangle inequality we obtain that:

${\begin{aligned}\quad \quad \mid (a_{n}-b_{n})-(A-B)\mid =\mid (a_{n}-A)+(B-b_{n})\mid \leq \mid a_{n}+A\mid +\mid B-b_{n}\mid <{\frac {\epsilon }{2}}+{\frac {\epsilon }{2}}=\epsilon \end{aligned}}$

• Therefore $\lim _{n\to \infty }(a_{n}-b_{n})=A-B$. $\blacksquare$

• Law 3 (Product Law of Convergent Sequences): If the limits of the sequences $\{a_{n}\}$ and $\{b_{n}\}$ are convergent, that is $\lim _{n\to \infty }a_{n}=A$ and $\lim _{n\to \infty }b_{n}=B$, then $\lim _{n\to \infty }(a_{n}b_{n})=\lim _{n\to \infty }a_{n}\cdot \lim _{n\to \infty }b_{n}=AB$.

• Proof of Law 3: Let $\{a_{n}\}$ and $\{b_{n}\}$ be convergent sequences. Since $\{a_{n}\}$ is convergent, it is also bounded, that is there exists $M\geq 0$ such that $\mid a_{n}\mid <M$. Now we note that $\lim _{n\to \infty }a_{n}=A$ implies that $\forall \epsilon >0\exists N_{1}\in \mathbb {N}$ such that if $n\geq N_{1}$ then $\mid a_{n}-A\mid <{\frac {\epsilon }{\mathrm {max} \{\mid b\mid ,1\}}}<\epsilon$. Similarly, $\lim _{n\to \infty }b_{n}=B$ implies that $\forall \epsilon >0\exists N_{2}\in \mathbb {N}$ such that if $n\geq N_{2}$ then $\mid b_{n}-B\mid <{\frac {\epsilon }{2\mathrm {max} \{M,1\}}}<\epsilon$. Note that these choices seem rather abstract, but will make more sense subsequently in the proof.

• Now we want to show that $\forall \epsilon >0\exists N\in \mathbb {N}$ such that if $n\geq N$ then $\mid a_{n}b_{n}-AB\mid <\epsilon$. Choose $N=\mathrm {max} \{N_{1},N_{2}\}$ so that both $\mid a_{n}-A\mid <{\frac {\epsilon }{\mathrm {max} \{\mid b\mid ,1\}}}$ and $\mid b_{n}-B\mid <{\frac {\epsilon }{2\mathrm {max} \{M,1\}}}$ hold. With some algebraic manipulation and the triangle inequality, we obtain that:

${\begin{aligned}\quad \mid a_{n}b_{n}-AB\mid =\mid a_{n}(b_{n}-B)+B(a_{n}-A)\mid \leq \mid a_{n}\mid \mid b_{n}-B\mid +\mid B\mid \mid a_{n}-A\mid \end{aligned}}$

• And making the appropriate substitutions we get that:

${\begin{aligned}\quad \mid a_{n}b_{n}-AB\mid \leq \mid a_{n}\mid \mid b_{n}-B\mid +\mid B\mid \mid a_{n}-A\mid <M{\frac {\epsilon }{2\mathrm {max} \{M,1\}}}+\mid b\mid {\frac {\epsilon }{2\mathrm {max} \{\mid b\mid ,1\}}}<{\frac {\epsilon }{2}}+{\frac {\epsilon }{2}}=\epsilon \end{aligned}}$

• Therefore $\lim _{n\to \infty }(a_{n}b_{n})=AB$. $\blacksquare$

• Law 4 (Quotient Law of Convergent Sequences): If the limits of the sequences $\{a_{n}\}$ and $\{b_{n}\}$ are convergent, that is $\lim _{n\to \infty }a_{n}=A$ and $\lim _{n\to \infty }b_{n}=B$, then $\lim _{n\to \infty }{\frac {a_{n}}{b_{n}}}={\frac {\lim _{n\to \infty }a_{n}}{\lim _{n\to \infty }b_{n}}}={\frac {A}{B}}$ provided that $\lim _{n\to \infty }b_{n}\neq 0$.

• Proof of Law 4: Let $\epsilon >0$ be given. We want to shown that $\forall \epsilon >0\exists N\in \mathbb {N}$ such that if $n\geq N$, then $\mid {\frac {a_{n}}{b_{n}}}-{\frac {A}{B}}\mid <\epsilon$. Doing some algebraic manipulation and using the triangle inequality, we get:

${\begin{aligned}\quad \quad {\bigg |}{\frac {a_{n}}{b_{n}}}-{\frac {A}{B}}{\bigg |}={\bigg |}{\frac {a_{n}}{b_{n}}}-{\frac {A}{b_{n}}}+{\frac {A}{b_{n}}}-{\frac {A}{B}}{\bigg |}={\bigg |}{\frac {a_{n}-A}{b_{n}}}+{\frac {A}{b_{n}}}-{\frac {A}{B}}{\bigg |}={\bigg |}{\frac {a_{n}-A}{b_{n}}}+(B-b_{n})\cdot {\frac {A}{B\cdot b_{n}}}{\bigg |}\leq {\bigg |}{\frac {a_{n}-A}{b_{n}}}{\bigg |}+{\bigg |}(B-b_{n})\cdot {\frac {A}{B\cdot b_{n}}}{\bigg |}\end{aligned}}$

• Doing some more work, we obtain that:

${\begin{aligned}\quad \quad {\bigg |}{\frac {a_{n}}{b_{n}}}-{\frac {A}{B}}{\bigg |}\leq {\frac {\mid a_{n}-A\mid }{\mid b_{n}\mid }}+{\bigg |}{\frac {A}{B}}{\bigg |}\cdot {\frac {\mid B-b_{n}\mid }{\mid b_{n}\mid }}\end{aligned}}$

• We note that since $\{b_{n}\}$ is convergent to the limit $B$, at some point $\mid b_{n}\mid >{\frac {\mid B\mid }{2}}$ which implies that ${\frac {1}{\mid b_{n}\mid }}<{\frac {2}{\mid B\mid }}$. Therefore:

${\begin{aligned}\quad \quad {\bigg |}{\frac {a_{n}}{b_{n}}}-{\frac {A}{B}}{\bigg |}\leq ...<{\frac {2}{\mid B\mid }}\mid a_{n}-A\mid +{\bigg |}{\frac {2A}{B^{2}}}{\bigg |}\cdot \mid B-b_{n}\mid \end{aligned}}$

• Now $\lim _{n\to \infty }a_{n}=A$ implies that $\forall \epsilon >0\exists N_{1}\in \mathbb {N}$ such that if $n\geq N_{1}$ then $\mid a_{n}+A\mid <{\frac {\mid B\mid \epsilon }{4}}$ to ensure that $\mid a_{n}-A\mid \cdot {\frac {2}{\mid B\mid }}<{\frac {\epsilon }{2}}$.

• Similarly, $\lim _{n\to \infty }b_{n}=B$ implies that $\forall \epsilon >0\exists N_{2}\in \mathbb {N}$ such that if $n\geq N_{2}$ then $\mid b_{n}+B\mid <{\bigg |}{\frac {B^{2}}{4A}}{\bigg |}\epsilon$ to ensure that $\mid b_{n}-B\mid {\bigg |}{\frac {2A}{B^{2}}}{\bigg |}<{\frac {\epsilon }{2}}$.

• Now choose $N=\mathrm {max} \{N_{1},N_{2}\}$ so that $\mid a_{n}+A\mid <{\frac {\mid B\mid \epsilon }{4}}<{\frac {\epsilon }{2}}$ and $\mid b_{n}+B\mid <{\bigg |}{\frac {2A}{B^{2}}}{\bigg |}\epsilon <{\frac {\epsilon }{2}}$, and therefore:

${\begin{aligned}\quad \quad {\bigg |}{\frac {a_{n}}{b_{n}}}-{\frac {A}{B}}{\bigg |}\leq ...<{\frac {2}{\mid B\mid }}\mid a_{n}-A\mid +{\bigg |}{\frac {2A}{B^{2}}}{\bigg |}\cdot \mid B-b_{n}\mid <{\frac {\epsilon }{2}}+{\frac {\epsilon }{2}}=\epsilon \end{aligned}}$

• Therefore for $n\geq N$, ${\bigg |}{\frac {a_{n}}{b_{n}}}-{\frac {A}{B}}{\bigg |}<\epsilon$, so $\lim _{n\to \infty }{\frac {a_{n}}{b_{n}}}={\frac {A}{B}}$. $\blacksquare$

• Law 5 (Constant Multiple Law of Convergent Sequences): If the limit of the sequence $\{a_{n}\}$ is convergent, that is $\lim _{n\to \infty }a_{n}=A$, and $k$ is a constant, then $\lim _{n\to \infty }ka_{n}=k\lim _{n\to \infty }a_{n}=kA$.

• Proof of Law 5:

• Let $k$ be a constant such that $k\neq 0$. We note that the case that $k=0$ is trivial since then $\lim _{n\ \to \infty }0a_{n}=\lim _{n\to \infty }0=0$. Let $\{a_{n}\}$ be a convergent sequence so that $\lim _{n\to \infty }a_{n}=A$ which implies that $\forall \epsilon \exists N_{1}\in \mathbb {N}$ such that if $n\geq N_{1}$ then $\mid a_{n}-A\mid <{\frac {\epsilon }{\mid k\mid }}$.

• Now let $\epsilon >0$ be given. We want to show $\lim _{n\to \infty }ka_{n}=kA$, that is, $\forall \epsilon >0\exists N\in \mathbb {N}$ such that if $n\geq N$ then $\mid k_{a}n-kA\mid <\epsilon$. Working with this inequality:

${\begin{aligned}\quad \mid ka_{n}-kA\mid =\mid k(a_{n}-A)\mid =\mid k\mid \mid a_{n}-A\mid <\epsilon \end{aligned}}$

• We choose $N=N_{1}$ and therefore if $n\geq N$ then $\mid a_{n}-A\mid <{\frac {\epsilon }{\mid k\mid }}$ and so $\lim _{n\to \infty }ka_{n}=kA$. $\blacksquare$

• Law 6 (Power Law of Convergent Sequences): If the limit of the sequence $\{a_{n}\}$ is convergent, that is $\lim _{n\to \infty }a_{n}=A$ and $k$ is a non-negative integer, then $\lim _{n\to \infty }(a_{n})^{k}=\left(\lim _{n\to \infty }a_{n}\right)^{k}=(A)^{k}$ provided that $k\in \mathbb {N}$.

• Proof of Law 6: Recall that from the power law for sequences that if $\{a_{n}\}$ and $\{b_{n}\}$ are convergent sequences such that $\lim _{n\to \infty }a_{n}=A$ and $\lim _{n\to \infty }b_{n}=B$, then $\lim _{n\to \infty }[a_{n}b_{n}]=AB$. The power law is just a special case of this.

• Consider $\{b_{n}\}=\{a_{n}\}$ and therefore by the power law $\lim _{n\to \infty }[a_{n}a_{n}]=\lim _{n\to \infty }(a_{n})^{2}=AA=A^{2}$. Therefore $\{a_{n}^{2}\}$ is a convergent sequence. Now continue to the process by letting $\{b_{n}\}=\{a_{n}\}^{2}$ and therefore $\lim _{n\to \infty }a_{n}a_{n}^{2}=AA^{2}=A^{3}$, etc… so we deduce if $\{a_{n}\}$ is convergent then $\lim _{n\to \infty }(a_{n})^{k}=A^{k}$.

• Law 7 (Squeeze Theorem for Convergent Sequences): If the limits of the sequences $\{a_{n}\}$, $\{b_{n}\}$, and $\{c_{n}\}$ are convergent and $a_{n}\leq b_{n}\leq c_{n}$ is true always after some $n^{\mathrm {th} }$ term, if $\lim _{n\to \infty }a_{n}=\lim _{n\to \infty }c_{n}=L$, then $\lim _{n\to \infty }b_{n}=L$.

• Proof of Law 7: Let $\{a_{n}\}$, $\{b_{n}\}$ and $\{c_{n}\}$ be convergent sequences such that $a_{n}\leq b_{n}\leq c_{n}$ is true always after some $M^{\mathrm {th} }$ term, and let $\lim _{n\to \infty }a_{n}=\lim _{n\to \infty }c_{n}=L$. We want to show that $\forall \epsilon >0\exists N\in \mathbb {N}$ such that if $n\geq N$ then >$\mid b_{n}-L\mid <\epsilon$.

• We note that $\lim _{n\to \infty }a_{n}=L$ implies that $\forall \epsilon >0\exists N_{1}$ such that if $n\geq N_{1}$ then $\mid a_{n}-L\mid <\epsilon$ or rather, $-\epsilon <a_{n}-L<\epsilon$.

• Similarly we note that $\lim _{n\to \infty }c_{n}=L$ implies that $\forall \epsilon >0\exists N_{2}$ such that if $n\geq N_{2}$ then $\mid c_{n}-L\mid <\epsilon$ or rather $-\epsilon <c_{n}-L<\epsilon$.

• Now let $N=\mathrm {max} \{M,N_{1},N_{2}\}$ to ensure that $-\epsilon <a_{n}-L<\epsilon$, $-\epsilon <c_{n}-L\epsilon$, and $a_{n}\leq b_{n}\leq c_{n}$. Subtracting $L$ from all parts of this inequality we get that $a_{n}-L\leq b_{n}-L\leq c_{n}-L$ or rather $-\epsilon <b_{n}-L<\epsilon$ so then $\mid b_{n}-L\mid <\epsilon$ and thus $\lim _{n\to \infty }b_{n}=L$. >$\blacksquare$

Example 1

Determine whether the sequence $\left\{{\frac {x^{2}-1}{x+1}}\right\}_{n=1}^{\infty }$ is convergent or divergent.

Let $f(n)={\frac {n^{2}-1}{n+1}}$ be a function analogous to our sequence. When we factor the numerator and cancel like-terms, we get that $\lim _{n\to \infty }f(n)=\lim _{n\to \infty }{\frac {n^{2}-1}{n+1}}=\lim _{n\to \infty }{\frac {(n+1)(n-1)}{n+1}}=\lim _{n\to \infty }n-1=\infty$. Therefore the sequence $\left\{{\frac {x^{2}-1}{x+1}}\right\}_{n=1}^{\infty }$ is divergent.

Licensing

Content obtained and/or adapted from:

• Limit of a Sequence, mathonline.wikidot.com under a CC BY-SA license
• Limit Sum/Difference Laws for Convergent Sequences, mathonline.wikidot.com under a CC BY-SA license
• Limit Product/Quotient Laws for Convergent Sequences, mathonline.wikidot.com under a CC BY-SA license
• Limit Constant Multiple/Power Laws for Convergent Sequences, mathonline.wikidot.com under a CC BY-SA license
• The Squeeze Theorem for Convergent Sequences, mathonline.wikidot.com under a CC BY-SA license